5

In Void Tyrant, when you or you opponent goes over the limit of the bar (overcharging is what I think it's called), they lose the round with a seemingly random amount less than what the winner has on there bar.

Is there any rhyme or reason to to what the enemies bar is set to when they go over, or is it just random?

2 Answers 2

5

Overcharging reduce the charge to 6, or 1 less than your opponent if they are 6 or less.

This also means that it doesn't matter how much overcharged you are, the amount of hits you get depends on your opponent's charge level when you overcharge. Therefore it is safer to attempt to gamble for a 12 when your opponent is 7 or below but if they are already at 9 or 10 it might be better to just hold.

This also applies to your opponent. If they overcharge while you are 6 or less you hit once, otherwise they are reduced to 6.

1
  • 1
    This answer seems to be in line with what you see on YouTube gameplay videos. Very important info, thanks!
    – jumxozizi
    Dec 30, 2020 at 9:50
3

Each point you go above 12 is subtracted to 12, with a maximum amount of 1 less than the opponent value. This is a confusing sentence, so, some examples:

  • You and your opponent are on 10 and you roll a 6. This means you overcharge by 4, so your final value is 12-4 = 8.
  • You are at 10, your opponent is at 5 and you roll a 6. Just like before, you overcharge by 4, so your final value is 12-4 = 8. This obviously doesn't make sense since you'd win with an overcharge, so your final value is 5-1 = 4.
  • I'm not sure what happens if you overcharge with your opponent at 0 since I never had that happen to me.

Hope that clears it up

1
  • 1
    On this gameplay the boss has 7 and draws a 6. Player has 9. By the logic in this answer, the boss should drop down to 8. On the video however you see the boss fall down to 6 instead.
    – jumxozizi
    Dec 30, 2020 at 9:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .